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As shown in the figure, the triangular pyramid a-bcd is cut by a plane, and the section is a parallelogram efgh
It is proved that the ∫ quadrilateral efgh is a parallelogram, ⊂ efgh, GH ⊂ plane BCD, ∩ plane BCD = CD, EF ⊂ plane ACD, ⊂ EF ∥ CD, and ef ⊂ plane efgh, CD ⊄ plane efgh, ∩ CD ∩ plane efgh
The triangular pyramid a-bcd is cut by a plane, and its cross section is parallelogram efgh
To prove by disprovement
Let CD not be parallel to efgh
CD does not belong to plane efgh
The CD must intersect the plane efgh
Let the intersection be m
CD belongs to plane ACD
Plane ACD intersects plane efgh at EF
According to axiom three, we can get m over EF
In the same way, we can get that M is on GH
So m is at the intersection of EF and GH
And ∵ EF ∥ GH has no intersection contrary to it
‖ CD ‖ plane efgh
What's wrong with this proof
Thank you, my friend, but it's not wrong here. Axiom 3 in our textbook is axiom 2 you said. Now we know the proof method. The counter proof method is not very good. I don't want to use it next time
This is a subject with serious defects. Where are e, F, G and H?
If the plane efgh passes the edge CD, the CD cannot be parallel to the plane efgh
The statement that plane ACD and plane efgh intersect in EF is too strong and arbitrary!
Of course, if the title shows EF through AC, ad, then this proof is correct
E. F.g.h is the center of the four sides of the regular triangular pyramid ABCD, and the volume ratio of the tetrahedral efgh to the tetrahedral ABCD is calculated?
If the side length of the regular triangular pyramid is one third of that of the original one, then the volume ratio is 1:27
It is known that the surface area of a tetrahedral ABCD is s, and the centers of its four faces are e, F, G, H. suppose the surface area of tetrahedral efgh is t, then t / s=——
Both the process and the answer are important
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